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en/6.2.11.md +2 −2
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$$E = \frac{\rho h}{\varepsilon_0}$$
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− Then, field generated by one of these biggest faces is the half of $E$. Let's consider now a point inside of one of these infinite plates. The resultant field over this point is the vectorial sum of the fields generated by lower negative plate, a plate with thickness $x$ (where $x$ is the distance from boundary between plates and the point) and a upper plate with thickness $h-x$:
+ Then, field generated by one of these biggest faces of the plate is the half of $E$. Let's consider now a point inside of one of these infinite plates. The resultant field over this point is the vectorial sum of the fields generated by lower negative plate, a plate with thickness $x$ (where $x$ is the distance from boundary between plates and the point) and a upper plate with thickness $h-x$:
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<p>
$$E_R(x) = \frac{\rho h}{2\varepsilon_0} + \frac{\rho (h-x)}{2\varepsilon_0} - \frac{\rho x}{2\varepsilon_0}$$
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<p>
$$E_R(x) = \frac{\rho}{\varepsilon_0}(h-x)$$
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− Hence, maximum field intensity is achieved when $x=0$, i.e., point is in boundary between plates.
+ Hence, maximum field intensity is achieved when $x=0$, i.e., the point is in the boundary between plates.
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<h4>Answer</h4>
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